Regular expression functions (regexp_match, regexp_replace) in PostgreSQL

Introduction

Regular expression functions help find or change text patterns inside data. They make searching and editing text easier and faster.

You want to find if a phone number matches a pattern like [phone].

You need to extract a part of an email address from a list of emails.

You want to replace all spaces in a text with underscores.

You want to clean up data by removing unwanted characters from strings.

Syntax

PostgreSQL

regexp_match(string, pattern [, flags])
regexp_replace(string, pattern, replacement [, flags])

regexp_match returns the first part of the string that matches the pattern as an array.

regexp_replace replaces parts of the string that match the pattern with the replacement text.

Examples

Finds the first group of digits in the string 'hello123'.

PostgreSQL

SELECT regexp_match('hello123', '\\d+');

Replaces spaces with underscores in the string 'hello world'.

PostgreSQL

SELECT regexp_replace('hello world', ' ', '_');

Extracts groups of letters and digits from the string.

PostgreSQL

SELECT regexp_match('abc123def', '([a-z]+)(\\d+)([a-z]+)');

Replaces digits in the string with '456'.

PostgreSQL

SELECT regexp_replace('abc123def', '\\d+', '456');

Sample Program

This example finds a phone number pattern in the text and then replaces it with a hidden version.

PostgreSQL

SELECT regexp_match('My phone is 555-1234', '\\d{3}-\\d{4}') AS phone_match;

SELECT regexp_replace('My phone is 555-1234', '\\d{3}-\\d{4}', 'XXX-XXXX') AS phone_hidden;

Important Notes

Remember to double backslashes (\\) in patterns because PostgreSQL uses backslash as an escape character.

Flags like 'i' can be added to make matching case-insensitive.

regexp_match returns NULL if no match is found.

Summary

Use regexp_match to find text patterns and get matching parts.

Use regexp_replace to change parts of text that match a pattern.

Regular expressions let you work with complex text rules easily.

Practice

1. What does the PostgreSQL function regexp_match('abc123', '\\d+') return?

easy

A. An error because of incorrect syntax

B. The string 'abc123' unchanged

C. An empty array because no digits are found

D. An array with the first sequence of digits found: ['123']

Solution

1. Step 1: Understand regexp_match function

   The function searches the input string for the first substring matching the pattern.

2. Step 2: Apply pattern '\\d+' to 'abc123'

   The pattern '\\d+' matches one or more digits. In 'abc123', '123' matches.

3. Final Answer:

   An array with the first sequence of digits found: ['123'] -> Option D

4. Quick Check:

   regexp_match returns first match array [OK]

Hint: regexp_match returns first matching group as array [OK]

Common Mistakes:

• Thinking regexp_match returns the whole string
• Expecting a string instead of an array
• Confusing regexp_match with regexp_replace

2. Which of the following is the correct syntax to replace all digits in 'abc123def' with '#' using regexp_replace in PostgreSQL?

easy

A. regexp_replace('abc123def', '\\d+', '#', 'g')

B. regexp_replace('abc123def', '[0-9]', '#', 'g')

C. regexp_replace('abc123def', '[0-9]+', '#')

D. regexp_replace('abc123def', '\\d', '#')

Solution

1. Step 1: Understand regexp_replace parameters

   regexp_replace(text, pattern, replacement, flags) replaces matching parts. To replace all matches, use flag 'g'.

2. Step 2: Choose pattern and flags to replace all digits

   Pattern '\\d+' matches one or more digits. Flag 'g' ensures all digit sequences replaced, not just first.

3. Final Answer:

   regexp_replace('abc123def', '\\d+', '#', 'g') -> Option A

4. Quick Check:

   Use 'g' flag to replace all matches [OK]

Hint: Use 'g' flag in regexp_replace to replace all matches [OK]

Common Mistakes:

• Omitting 'g' flag replaces only first match
• Using '\\d' without 'g' replaces only one digit
• Confusing pattern syntax with wrong brackets

3. What is the result of this query?

SELECT regexp_match('user_2023_data', 'user_(\\d{4})_data');

medium

A. NULL

B. ['user_2023_data']

C. ['2023']

D. An error due to wrong pattern

Solution

1. Step 1: Analyze the pattern 'user_(\\d{4})_data'

   The pattern looks for 'user_' followed by exactly 4 digits (captured), then '_data'.

2. Step 2: Match pattern against 'user_2023_data'

   The string matches perfectly, capturing '2023' as the group.

3. Final Answer:

   ['2023'] -> Option C

4. Quick Check:

   Captured group returns ['2023'] [OK]

Hint: Captured groups in regexp_match return as array [OK]

Common Mistakes:

• Expecting full string instead of captured group
• Not escaping backslashes properly
• Confusing regexp_match with regexp_replace output

4. Identify the error in this query:

SELECT regexp_replace('abc123def', '\\d+', '#');

It is intended to replace all digit sequences with '#'.

medium

A. regexp_replace cannot replace digits

B. Missing 'g' flag, so only first digit sequence replaced

C. Replacement '#' must be in double quotes

D. Pattern '\\d+' is invalid syntax

Solution

1. Step 1: Understand regexp_replace behavior without flags

   Without 'g' flag, regexp_replace replaces only the first match.

2. Step 2: Check the pattern and replacement

   Pattern '\\d+' is valid and replacement '#' is correct. Only missing 'g' flag causes partial replacement.

3. Final Answer:

   Missing 'g' flag, so only first digit sequence replaced -> Option B

4. Quick Check:

   Use 'g' flag to replace all matches [OK]

Hint: Add 'g' flag to replace all matches in regexp_replace [OK]

Common Mistakes:

• Assuming all matches replaced without 'g'
• Thinking pattern syntax is wrong
• Using single quotes incorrectly for replacement

5. You want to extract the domain name from email addresses like '[e-mail]' using regexp_match. Which query correctly extracts 'example'?

hard

A. SELECT (regexp_match('[e-mail]', '@([\\w]+)\\.com'))[1];

B. SELECT regexp_match('[e-mail]', '@([\\w]+)\\.com');

C. SELECT regexp_replace('[e-mail]', '.*@([\\w]+)\\.com', '\\1');

D. SELECT regexp_replace('[e-mail]', '@([\\w]+)\\.com', '\\1');

Solution

1. Step 1: Understand regexp_match returns array of groups

   regexp_match returns an array with captured groups. To get the domain, access first element [1].

2. Step 2: Check pattern '@([\\w]+)\\.com'

   This pattern captures word characters after '@' and before '.com', which is the domain.

3. Final Answer:

   SELECT (regexp_match('[e-mail]', '@([\\w]+)\\.com'))[1]; -> Option A

4. Quick Check:

   Access first group from regexp_match array [OK]

Hint: Use (regexp_match(...))[1] to get captured group string [OK]

Common Mistakes:

• Not accessing array element from regexp_match
• Using regexp_replace without global flag incorrectly
• Confusing pattern syntax for domain extraction